The price p and the quantity x sold of a certain product obey the demand equatio

Question

The price p and the quantity x sold of a certain product obey the demand equation p = -1/3x + 70, 0 x 210 Express the revenue R as a function of x. What is the revenue if 69 units are sold? What quantity x maximizes revenue? What is the maximum revenue? Express the revenue R as a function of x. R(x) = What is the revenue if 69 units are sold? R = $ (simplify your answer.) What quantity x maximizes revenue? x = units(Simplify your answer.) What is the maximum revenue? R = $ (Simplify your answer.) What price p should the company charge to maximum revenue? What price p should the company charge to maximum revenue? p = $ (Simplify your answer.)

Explanation / Answer

a)revenue= price *quantity=x*(-1/3x+70)=-1/3x^2+70x..........b)R(69)=69*(-1/3(69)+70)=3243.......c) differentiating R(x) and equating it to 0 we get x=105, hence the revenue is max at x=105,.......d) price at x=105 is p=-1/3(105)+70=35.............plz rate :)

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